Lightoj 1003 - Drunk(拓扑排序)

One of my friends is always drunk. So, sometimes I get a bit confused whether he is drunk or not. So, one day I was talking to him, about his drinks! He began to describe his way of drinking. So, let me share his ideas a bit. I am expressing in my words.

There are many kinds of drinks, which he used to take. But there are some rules; there are some drinks that have some pre requisites. Suppose if you want to take wine, you should have taken soda, water before it. That's why to get real drunk is not that easy.

Now given the name of some drinks! And the prerequisites of the drinks, you have to say that whether it's possible to get drunk or not. To get drunk, a person should take all the drinks.

Input

Input starts with an integer T (≤ 50), denoting the number of test cases.

Each case starts with an integer m (1 ≤ m ≤ 10000). Each of the next m lines will contain two names each in the format a b, denoting that you must have a before having b. The names will contain at most 10 characters with no blanks.

Output

For each case, print the case number and 'Yes' or 'No', depending on whether it's possible to get drunk or not.

Sample Input	Output for Sample Input
2 2 soda wine water wine 3 soda wine water wine wine water	Case 1: Yes Case 2: No

拓扑排序，判断是否有环。

嗯...只要剩下点就是有环 好久没看自己以前 写的博客了http://fengweiding.blog.163.com/blog/static/2300541212014112831155847/

/* ***********************************************
Author        :guanjun
Created Time  :2016/6/7 20:22:01
File Name     :1003.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
    int x,y;
};
struct cmp{
    bool operator()(Node a,Node b){
        if(a.x==b.x) return a.y> b.y;
        return a.x>b.x;
    }
};

bool cmp(int a,int b){
    return a>b;
}
vector<int>edge[maxn];
int m,num;
map<string,int>mp;
int in[maxn];
bool topsort(){
    int cnt=0;
    queue<int>q;
    for(int i=1;i<=num;i++)if(in[i]==0)q.push(i);
    while(!q.empty()){
        int u=q.front();q.pop();
        cnt++;
        for(int i=0;i<edge[u].size();i++){
            int v=edge[u][i];
            in[v]--;
            if(in[v]==0){
                q.push(v);    
            }
        }
    }
    if(cnt<num)return false;
    return true;
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int T;
    string s1,s2;
    cin>>T;
    for(int t=1;t<=T;t++){
        cin>>m;
        num=0;mp.clear();cle(in);
        for(int i=0;i<maxn;i++)edge[i].clear();
        for(int i=1;i<=m;i++){
            cin>>s1>>s2;
            if(!mp[s1])mp[s1]=++num;
            if(!mp[s2])mp[s2]=++num;
            edge[mp[s1]].push_back(mp[s2]);
            in[mp[s2]]++;
        }
        printf("Case %d: %s\n",t,topsort()?"Yes":"No");
    }
    return 0;
}